\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}-\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)\) (do \(\sqrt{2}+1,\sqrt{2}-1>0\))
\(=\sqrt{2}+1-\sqrt{2}+1\)
\(=2\)
Vậy \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=2\)